Two different wires having lengths L1 and L2 and respective temeperature coefficient of linear expansion -1 and -2 are joined end to end- Then the effective temperature coefficient of linear expansion is-

Let nbsp;L'_1 and L'_2 be the lengths of the wire when temperature is changed by Δ T^∘ C At T^∘ C L_eq=L_1+L_2 At T+Δ ^∘ C L'_eq=L'_1+L'_2 ∴ L_eq(1+α _eqΔ T) ...

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Standard XII

Physics

Scale Conversions

Two different...

Question

Two different wires having lengths $L_{1}$ and $L_{2}$ and respective temeperature coefficient of linear expansion $α_{1}$ and $α_{2}$ are joined end to end. Then the effective temperature coefficient of linear expansion is:

\frac{α L_{1} + α_{2} L_{2}}{L_{1} + L_{2}}

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2 \sqrt{α_{1} α_{2}}

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\frac{α_{1} + α_{2}}{2}

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\frac{α_{1} α_{2}}{α_{1} + α_{2}} \frac{L_{2} L_{1}}{(L_{2} + L_{1})^{2}}

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Solution

The correct option is A $\frac{α L_{1} + α_{2} L_{2}}{L_{1} + L_{2}}$
Let $L_{1}^{'}$ and $L_{2}^{'}$ be the lengths of the wire when temperature is changed by $Δ T^{\circ} C$
At $T^{\circ} C$
$L_{e q} = L_{1} + L_{2}$
$At T + Δ^{\circ} C$
$L_{e q}^{'} = L_{1}^{'} + L_{2}^{'}$
$∴ L_{e q} (1 + α_{e q} Δ T) = L_{1} (1 + α_{1} Δ T) + L_{2} (1 + α_{2} Δ T) [∵ L^{'} = L (1 + α Δ T)]$
$\Rightarrow (L_{1} + L_{2}) (1 + α_{e q} Δ T) = L_{1} + L_{2} + L_{1} α_{1} Δ T + L_{2} α_{2} Δ T$
$\Rightarrow α_{e q} = \frac{α_{1} L_{1} + α_{2} L_{2}}{L_{1} + L_{2}}$

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Two different wires having lengths L1 and L2 and respective temeperature coefficient of linear expansion α1 and α2 are joined end to end. Then the effective temperature coefficient of linear expansion is:

Two different wires having lengths $L_{1}$ and $L_{2}$ and respective temeperature coefficient of linear expansion $α_{1}$ and $α_{2}$ are joined end to end. Then the effective temperature coefficient of linear expansion is: